Seeing as one can expand any function on the sphere in terms of the spherical harmonics, I was thinking it should be possible to express the function for a sphere itself in terms of them. I have started trying to do this, by calculating the first five expansion coefficients according to the standard formula:
$$ f_{lm}=\int_0^{2\pi}d\phi\int_0^\pi \sin(\theta)\,d\theta f(\theta,\phi) Y_{lm}^*(\theta,\phi) $$
where $$ f(\theta,\phi)=\sqrt{1-\sin^2(\theta)\cos^2(\phi)-\sin^2(\theta)\sin^2(\phi)} $$
is the function for a sphere in spherical coordinates. My confusion arises because the only nonzero coefficient I calculated was $f_{00}=\sqrt{\pi}$, meaning I have the following expansion:
$$ f(\theta,\phi)=\sqrt{\pi}Y_{00}\\ =\frac{1}{2}\,. $$
This seems really strange to me, to have an expansion in terms of a constant. Is my problem that I haven't looked at high enough expansion co-efficients? I have only calculated $f_{00},f_{1,0},f_{1,1},f_{1,-1},f_{2,-2}$. Or is it a more fundamental misunderstanding on my behalf?
Any help kindly appreciated!
This is an answer to the question asked in a comment on my previous answer, which seems to be at the core of what OP really wants. That question is " I just want a function for the sphere (in spherical coordinates so I can expand it in terms of the spherical harmonics). Is my problem that this is an ill-defined question?"
When you say "I want a function for ", you need to be pretty precise. Let me talk about the circle for a moment, and three possible "functions for a circle." First, there's
$$ F(x, y) = x^2 + y^2 - 1. $$ or more formally, $$ F: \mathbb R^2 \to R : (x, y) \mapsto x^2 + y^2 - 1. $$ This function has the property that it evaluates to zero exactly on the circle. If $F(a, b) = 0$, then $(a, b)$ is a circle point, and vice versa. It's called "an implicit representation of the circle". You could also write down
$$ G(x, y) = (x^2 + y^2 - 1)^2, \text{or} \\ H(x, y) = \arctan(x^2 + y^2 - 1) $$ which are other implicit representations of the circle. So you can't talk about "the" implicit representation.
Here's another function: $$ u : [0, 2\pi) \to \mathbb R^2: t \mapsto (\cos t, \sin t) $$
This, too, describes a circle; it's called a "parameterization" of the circle, and has the property that for each $t$, $u(t)$ is a point of the circle, and each point of the circle is $u(t)$ for some $t$. Usually those conditions are weakened a little, and it's considered OK for a parameterization to cover the circle multiple times (as would happen if you enlarged the domain of $u$ to the whole real line), or to hit a few points more than once (as would happen if you made the domain of $u$ into $[0, 2 \pi]$), or even to miss a point or two, so that the make a "real" parameterization, you'd need several functions whose images overlapped. As an example, consider $$ v: \mathbb R \to \mathbb R^2 : t \mapsto (\frac{2t}{1+t^2}, \frac{1 - t^2}{1 + t^2}), $$ which nicely parameterizes all but one point of the circle, namely $(0, -1)$.
There's no single parameterization, however; you can't talk of the parameterization of the circle.
And then there's a third notion: that the circle looks locally like the graph of a function. So you write a function $$ f: [-1, 1] \to \mathbb R: x \mapsto \sqrt{1 - x^2}. $$ That has the property that for every $x$, you have $(x, f(x))$ is a point of the circle, but to get the whole circle, you'd need to also consider a second function, $$ g: [-1, 1] \to \mathbb R: x \mapsto -\sqrt{1 - x^2}. $$
One thing to notice about all these descriptions is their domains. The first, $F$ is defined on all of $R^2$; $u$ is defined on an interval. So are $f$ and $g$. Not one of them is defined on a circle, i.e., not one of them is suitable for expansion in the Fourier basis for $L^2$ functions on $S^1$.
An exact analogy holds for $S^2$: spherical harmonics are a tool for expressing functions from $S^2$ to $\mathbb R$ nicely. But the domain has to be the sphere.
Let me write down one last way to describe a circle: I'm going to parameterize it with a circle:
$$ q : S^1 \mapsto \mathbb R^2 : \theta \mapsto (\cos \theta, \sin \theta) $$
Now each coordinate of $q$, i.e., $$ q_1 : S^1 \mapsto \mathbb R^2 : \theta \mapsto \cos \theta\\ q_2 : S^1 \mapsto \mathbb R^2 : \theta \mapsto \sin \theta $$ is actually a function on $S^1$, and amenable to expansion in the Fourier basis $$ 1, \cos \theta, \sin \theta, c\cos 2\theta, c\sin 2\theta, \ldots $$ The first has coefficients $(0, 1, 0, 0, \ldots)$, the second has coefficients $(0, 0, 1, 0, 0, \ldots)$.
Now to formulate your question about expressing "a function for the sphere" in spherical harmonics, you need to decide what function you're talking about, and you need to be sure that its domain is in fact the unit sphere. Once you've done that, you can write down the function, and others may be able to help you evaluate its coefficients in the harmonic basis. It cannot, however, be defined on half of the sphere, or defined on a disk, or a rectangle, etc.; asking for an expansion of such a function in the harmonic basis would be like asking for an expansion of the vector $[0.1, 1.5, -4] \in \mathbb R^3$ in terms of the standard basis for "the polynomials of degree five or less" -- it really doesn't even make sense.
ADDED AFTER COMMENTS: If you want to do something analogous to $q$ for the unit sphere, you need to choose some way to express $q$. For the unit circle, there's the convenient $\theta$ parameterization, which I used. But I could instead have defined $$ \bar{q} : S^1 \to \mathbb R^2 : (x, y) \mapsto (x, y) $$ where now $\bar{q}$ is defined in terms of the $x$- and $y$-coordinates of each point. Note that as functions on $S^1$, these functions $q$ and $\bar{q}$ are really the same: the rightmost point is sent to $(1, 0)$, the topmost point to $(0, 1)$, etc.
If you want to express $\bar{q}_1 (x, y) = x$ in the Fourier basis, you need to express the Fourier basis in $xy$-coordinates as well. It turns out that the functions we usually write as $$ 1, \cos \theta, \sin \theta, c\cos 2 \theta, c \sin 2\theta, \ldots, $$ in $xy$-coordinates, are just $$ 1, x, y, c(x^2 - y^2), c(2xy), \ldots $$
For an analogous function defined on the sphere, say
$$ Q: S^2 \to \mathbb R^3, $$ we must once again choose coordinates. If you like the $(\theta, \phi)$ representation for coordinates on $S^2$, but $xyz$ on $\mathbb R^3$, then $$ Q(\theta, \phi) = (\cos \theta \sin \phi, \sin \theta \sin \phi, cos \phi), $$ where here $\phi$ goes from $0 to $\pi$. (I hope that's the convention that you're using!)
The first coordinate function, $$ Q_1(\theta, \phi) = \cos \theta \sin \phi, $$ is a real-valued function on the sphere, and can therefore be expressed in harmonics. In fact...it's exactly equal to one of the standard harmonics. The same goes for $Q_2$ and $Q_3$.
Alternative, you could define $$ \bar{Q}(x, y, z) = (x, y, z), $$ in which case the first coordinate function is just $(x, y, z) \mapsto x$. Since the harmonics, expressed in $xyz$ coordinates, are just (up to scalar multiples) $$ 1, x, y, z, x^2 - y^2, 2xy, \ldots $$ we once again see that this coordinate function $\bar{Q}_1$ is exactly equal to one of the harmonics. This is what the first few paragraphs of my other answer were getting at.